Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#define ll long long
using namespace std;
const int N=510,MX=1e7;
int p[250010];
int vis[MX];
ll rec[N][N];
int n,m,cnt;
void prework(int n){
	vis[1]=false;
	int cnt=0;
	for (int i=2;cnt<n;++i){
		if (!vis[i])
			p[++cnt]=i;
		for (int j=1;j<=cnt&&i*p[j]<MX;++j){
			vis[i*p[j]]=1;
			if (i%p[j]==0) break;
		}
	}
}
ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
ll lcm(ll x,ll y){
	if (!x||!y) return x+y;
	return x/gcd(x,y)*y;
}
void solve(){
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j)
			if ((i+j)%2==0)
				rec[i][j]=1LL*p[(i+j)/2]*p[n+(i-j)/2+(n+1)/2];
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j)
			if (!rec[i][j])
				rec[i][j]=lcm(lcm(rec[i-1][j],rec[i][j+1]),lcm(rec[i+1][j],rec[i][j-1]))+1;
}
 
int main(){
#ifndef ONLINE_JUDGE
	//freopen("a.in","r",stdin);
#endif
	ll tmp;
	scanf("%d",&n);
	if (n==2){
		printf("%d %d\n%d %d\n",4,7,23,10);
		return 0;
	}
	prework(n*n);
	solve();
	for (int i=1;i<=n;++i){
		for (int j=1;j<=n;++j)
			printf("%lld ",rec[i][j]);
		printf("\n");
	}
}